CHANAKYA ACADEMY OF PROFESSIONAL STUDIES Jr|11th IIT...

1THERMODYNAMICS

CHANAKYA ACADEMY OF PROFESSIONAL STUDIES

Jr|11th IIT-JEE-MAIN|NEET|CHEMISTRY:VOL-II

2 THERMODYNAMICS



Introduction and Concepts Thermodynamics The subject dealing with Quantitative relation

between heat energy and other forms of energy inphysico - Chemical processes is calledThermodynamics.

Chemical Thermodynamics The branch of thermodynamics which deals with

the study of process in which chemical energy isinvolved is called chemical thermodynamics.

These results are formulated into four law’s namelyZero, First , second and third laws ofthermodynamics.

These laws are based on experimental facts butnot on the theoretical facts.

Thermodynamics predicts the energytransformations and feasibility of a process.

Thermodynamics deals heat changes occuringbetween system and surroundings.Types of rection

Chemical change reactions;

Eg: 3 2s s gCaCO CaO CO

Physico change reactions

Eg: 2 2s lH O H O

Allotropic change reactions

Eg: graphite diamondC CPhase : It is homegenious portion which hasidentical physical & chemical properties throughout it & physically distinctable & mechanicalseparable from other portions.

Both Physico change & Allotropic change reactionsare phase change reactions.

In solid mixture, number of phases is equal tonumber of individual solids.

Eg: graphite diamondC C , number of phases isequal to two.

Thermodynamics deals with the calculation of Q,E, E , H , S & G in all type of reactionsand gas in cylinder system based on all alws ofthermodynamics.The terms used in thermodynamics:System:

It is any part of universe that is underthermodynamic study at that instant.Eg: a) A crystal (for a crystallographer)b) A physical process (for a physicist)c) Chemical reaction (for a chemist)

Surroundings The remaining part of the universe, other than system

is called surroundings.Universe = system + surroundings

Surroundings

System

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Boundary: Any real (or) imaginary, rigid (or) non-rigid surface that separates system and surroundings.

Diathermic wall : which allows heat exchangingbetween system and surroundings.Adiabatic wall : which does not allow heatexchanging between system and surroundings.

Types of system: Systems are classified on the basis of their

interaction with the surroundings as follows:1. Open System : The system where matter andenergy are exchanged with surroundings. Boundaryis not sealed and not insulated.

Surroundings

System

(Heat + Work)

Energy

(A) Open System

Eg: All living beings, Reactants in open vessel.2. Closed system: The system where only theenergy but not the matter is exchanged with thesurroundings. Boundary is sealed but not insulated.

Surroundings

System

Matter

Energy

(B) Closed System

Eg: A closed steel container having hot water.3. Isolated system: The system which does notexchange either the matter or energy with thesurroudings. Boundary is sealed and insulated

Matter

Energy

(C) Isolated System

Eg: A perfectly insulated, closed flask containing water. Universe is treated as an isolated system. On the basis of composition, there are two types

of systems.1) Homogeneous system : A system consistingof one phase.Eg: Pure solid, a liquid or a mixture of gases.2) Heterogeneous system : A system consistingof two or more phases.Eg: a solid in contact with liquid state, solid in solidmixtureEg: water at triple point consists three phases.

Solid Liquid

SupercriticalRegion

Triple Point

Critical Point(TC, Pc)

Gas

Pres

s ure

(Mp a

)

Temperature (K)

On the basis of number of particles,there are twotypes of systems.1) Micro system : Which consists less number ofconstituent particles.2) Macro system : Which consists huge numberof constituent particles.

Any system in thermodynamics is macro scopicsystem.

Thermodynamics deal with energy changes ofmacroscopic systems involving a large number ofmolecules rather than microscopic systemsconatining a few molecules. Thermodynamics is notconverned above how and at what rate these energytransformation are carred out, but is based on inital

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and final states of a system undergoing the change.Laws of thermodynamics apply only when a systemis in equilibrium or moves from one equilibrium stateto another equilibrium state.Limitations of Thermodynamics

These laws donot give any idea about the rates ofthe processes.

It fails to explain the systems away from equilibrium.

State of a system Macro scopic properties : These are the

properties which are associated with the macroscopic system.Eg: P, V, T, n, G, H, E etc

The system is said to be in a certain state, when it’smacroscopic properties have definite values. It isdefined in terms of its state functions such as P, V,T etc.

If any one of the state functions is changed, thestate of that system is said to be changed.Eg: at 1atm pressure H2O isa) solid below 00Cb) liquid between 00C - 1000Candc) gas above 1000C.

State functions or state variables The Thermodynamic properties whose values

depend only upon the initial and final states of thesystem and are independent of the path are calledstate functions.Eg: Internal energy (E), Enthalpy (H),Entropy (S), Gibb’s energy (G), Pressure (P),Temperature (T), Volume (V), Helmholtz energy(A) reversible isothermal work done, chamicalpotential ( )Path Function:

The property of a system that depends on the pathof the process.Eg: work, heat.

Internal energy (E), enthalphy (H), entropy (S),Gibb’s energy (G), Helmholtz energy (A0 are thedirect froms of energies of the system.

Heat (Q) and work done are the indirect form ofenergies and appread at the boundary of the system.Thermodynamic Process:

The operation which brings about the changes inthe state of the system is termed as thermodynamicprocess.Isothermal Process :

A process in which temperature of system doesnot change throughout the studies. For an isothermalprocess dT=0 and dU=0. An isothermal processis achieved by using thermostatic control.Adiabatic Process:

A process in which exchange of heat betweensystem and surroundings does not take place.For an adiabatic process q = 0.It can be achieved by insulating the boundaries ofsystem.Isobaric Process:

A process in which pressure of the system remainsconstant throughout the studies.For an isobaric process 0P .Isochoric Process:

A process in which volume of the system remainsconstant throughout the studies.

For an isochoric process 0V .Cyclic Process:

A process in which initial state of system is regainedafter a series of operations.For a cyclic process 0U and 0H ,... etc. (State function) = 0, (path function) 0

Thermodynamic Equilibrium : When a system exists simultaneously in a thermal

equilibrium, mechanical equilibrium and chemicalequilibrium, is said to be in thermodynamicequilibrium.

Equilibrium Condition

Thermal Equilibrium

T=0 i.e constant temperature

Chemical Equilibrium

C=0 i.e constant concentration

Mechanical Equilibrium

P=0, F=0, W=0 i.e constant pressure

Reversible Process: A reversible process (or quasistatic process) is one in

which all changes occuring at any part of the processare exactly reversed when change is carried out inopposite direction. It gives rise to maximum work.

Driving force is Infinitesimal greater than that of

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opposing force. In reversible expansion of gas in cylinder system,

gas externalP P dP

In reversible compression of gas in cyclindersystem, gas externalP P dP

dP = Infinitesimal change pressure It is unreal process Which is achieved by considering ideal gas is placed

in cylinder fitted with weight less and frictionlesspiston.Irreversible Process:

An irreversible process is one in which direction ofthe change cannot be reversed by small change invariables.An irreversible process is a real one and all processwhich naturally occur are irreversible.It involvesa) A fast change during investigationb) Driving force is much higher than opposing force

c) . . Expansionrev irrevW W

d) . . Compressionrev irrevW W Graphical representation of thermodynamic

processes

P

V

(4) (3)

(2)

(1)

WIsobaric > WIsothermal > WAdiabatic > WIsochoric1) Isobaric process 2) Isothermal process3) Adiabatic process 4) Isochoric process

Extensive Property It is the property of a substance that depends on

the quantity or size of matter present in the system.Eg: Mass, volume of a gas, Internal energy,Enthalpy, entropy, heat capacity, Gibbs energy, heatcontent, no of moles etc.

Intensive Property It is the property of a substance that does not

depend on the quantity or size of matter present inthe system.Eg: Density, molar properties (such as molarvolume, molar entropy,molar heat capacity) surface

tension, viscosity, specific heat, refractiveindex,pressure,temperature, boiling point,freezingpoint, vapour pressure etc.,

Extensive property is additive property Intensive property is non-aditive property.

x=extensive property y=extensive property

x=extensive property y=intensive property

x=intensive property y=intensive property

.x y Ext E Q W x+y=Ext. x+y=Int.

x-y=Ext.(Q= E -W) x-y=Ext. x-y=Int.

.x MInt dy V

.x Exty .x Int

y

. ,dx HInt T S H Tdy S

xy=Ext. xy=Int.

x.y=Ext

Comparsion of isothermal and adiabaticprocess till same final pressure

P

v

1P Isothermal

Adiabatic

adiV

fP

isoV

iii)Adiabatic process

Workdone

Isotherms

V

P

Introduction and Types Of Systems1. Thermodynamic laws speak about

1) rates of chemical changes2) feasibility and energy transformations of aprocess3) Both the rate and energy changes of a process4) Energy changes in chemical reactions only

2. The object under thermodynamical study iscalled1) System 2) Universe3) Surrounding 4) Boundary

3. Which of the following are true about a

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“system”1) Will do not have definite amount of substance2) Energy and matter may not be exchanged withsurroundings3) (Universe+surroundings)4) (Universe-surroundings)

4. Hot water in a thermos flask is an examplefor1) Isolated system 2) Open system3) Closed system 4) Adiabatic system

5. In open system, system and surroundingsexchange1) Energy only 2) Matter only3) Energy and matter4) Neither energy nor matter

6. In a closed system1) Energy is not exchanged2) Matter is exchanged3) Energy is only exchanged4) Energy and matter are exchanged

7. “Closed system” is1) Perfectly sealed 2) Perfectly insulated3) Both 1 & 24) Neither insulated nor sealed

State Functions and Path Functions:8. Which one of the following statement is false

1) Work is a state function2) Temperature is a state function3) Change in the state is completely defined whenthe initial and final states are specified4) Work appears at the boundary of the system

9. Which of the following is a path function1) Internal energy 2) Enthalpy3) Work 4) Entropy

10. Which of the following statement is correct ?1) Only internal energy is a state function but notwork2) Only work is a state function but not internalenergy3) Both internal energy and work are statefunctions4) Neither internal energy nor work is a statefunction

Thermo Dynamical Processes11. A process in which no heat change takes place

is called1) An isothermal process 2) An adiabatic process

3) An isobaric process 4) An isochoric process12. A gaseous system changes from state

A(P1,V1,T1) to B (P2,V2,T2), B to C (P3,V3T3)and finally from C to A. The whole processmay be called1) Cyclic process 2) Reversible process3) Isobaric process 4)Spontaneous process

13. For a cyclic process, the condition is1) 0U 2) 0H 3) 0U and 0H 4) both 0U and 0H

14. An adiabatic expansion of an Ideal gas alwayshas1) Constant in Temperature 2) q=03) w=0 4) 0H

01) 2 02) 1 03) 4 04) 1 05) 3 06) 307) 1 08) 1 9) 3 10) 1 11) 2 12) 113) 4 14) 2

Thermo Dynamical Processes:1. Which is an irreversible process ?

A) Mixing of two gases by diffusionB) Evaporation of water at 373K and 1atmpressureC) Dissolution of NaCl in waterD) Burning of CoalThe correct answer1) A,B 2) B,C 3) A,B,C 4) A,C,D

2. 0U for1) Cyclic process, Adiabatic process2) Isothermal, Adiabatic process3) Cyclic process, Isothermal process4) Isochoric process, Isothermal process

Internal Energy, Enthalpy and Work done :3. The work done when a gas is compressed by

an average pressure of 0.50 atm so as todecrease its volume from 400 cm3 to 200cm3

1) 10.13 J 2) 20.13J 3) 30.13 J 4) 40.13 J4. Temperature of 1 mol of a gas is increased

by 1º at constant pressure. Work done is :

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1) R 2) 2 R 3) R/2 4) 3 R5. A certain electric motor produced 15 KJ of

energy each second as numerical work and lost2KJ as heat to the surroundings . What is thechange in the internal energy of the motor andits power supply each second.1) -17 KJ 2) -17 J 3) 27 KJ 4) 23 KJ

6. Two liters of N2 at O0C and 5atm are expandedexothermally against a constant externalpressure of 1atm untill the pressure of gasreaches 1atm. Assuring gas to be ideal.Calculate work of expansion1) 400 J 2) -810 J 3) -810 KJ 4) 600 KJ

7. When work done by a system was 10J , theincrease in the internal energy of the systemwas 30J. The heat ‘q’ supplied to the systemwas:1) -40J 2) +20J 3) 40J 4) -20J

8. The work done when 6.5g of zinc reacts withdil HCl is an open beaker at 298 K is1) -495.52 J 2) 247.76 J3) -247.76 J 4) -123.88 J

9. The work done in heating one mole of an idealgas at constant pressure from 015 C to 025 C is1) 1.987 cal 2) 198.7 cal3) 9.935 cal 4) 19.87 cal

10. A gas expands from 1.5 to 6.5 L against aconstant pressure of 0.5 atm and during thisprocess the gas also absorbs 100 J of heat. Thechange in the internal energy of the gas is1) 153.3J 2) 353.3J3) –153.3J 4) -353.3J

11. 10 g of argon gas is compressed isothermallyand reversibly at a temperature of 27º C from10 L to 5 L. Calculate q and H for thisprocess. Atomic wt. of Ar = 40.

1) + 103.99 cal, 0 2) + 39.99 cal, 3 J3) - 39.99 cal, 3 J 4) -103.99 cal, 0

First Law of Thermo dynamics :12. A system absorbs 20 KJ of heat and also does

10KJ of work. The net internal energy of thesystem1) Increases by 10 KJ 2) Decreases by 10 KJ3) Increases by 30 KJ 4) Decreases by 30 KJ

13. If a gas absorbs 200 J of heat and expands by500 cm3 against a constant pressure of 2 x 105

Nm–2, then change in internal energy is

1) -300 J 2) -100 J 3) +100J 4)+300 J14. In an insulated container 1 mole of a liquid.

molar volume 100 ml at 1 bar. Liquid is steeplytaken to 100 bar, when volume of liquiddecreases by 1 ml. Find H for the process.1) 7900 bar ml 2) 8900 bar ml3) 9900 bar ml 4) 10900 bar ml

01) 4 02) 3 03) 1 04) 1 05) 1 06) 207) 3 08) 3 09) 4 10) 3 11) 1 12) 113) 3 14) 3

1. All natural processes are thermodynamicallyirreversible.

2. 0U for isothermal and cyclic processes3. W P V 4. PV = RT at TK

1P V V R T at 1T K

P V R 5. Q U W

6. 2 1W P V V 7. Q U W

8. 2 22S aq aq gZn HCl ZnCl H

9. Rise of 010 1.987 10 19.87C cal 10. Q U W

11. 2 1W P V V H U P V

12. U Q W

13. U Q W U Q W 14. Volume of 1 mole liquid

= 100 ml at pressure one bar= 99 ml at pressure 100 bar

2 1W P V V H U P V

8 THERMODYNAMICS



Internal Energy, Enthalpy and Work Done,First Law

1. The workdone in ergs for the reversibleexpansion of one mole of an ideal gas from avolume of 10 litres to 20 litres at 250C is1) 2.303 298 0.082 log22) –298 107 8.31 2.303log23) 2.303 298 0.082 log 0.54) –2.303 298 2 log2

2. An ideal gas expands from 10-3m3 to 10-2m3 at300K against a constant pressure of 105Nm-2.The workdone is1) -103 kJ 2) 102kJ 3) -0.9 kJ 4) -900 kJ

3. A system has internal energy equal to E1,450J of heat is taken out of it and 600J ofwork is done on it. The final energy of thesystem will be1) (E1+150) 2) (E1+1050)3) (E1-150) 4) None of these

4. If the internal energy of 22g CO2 at 273K is‘U’, internal energy of which of the followingis ‘4U’ at same T ?1) 5.5 CO2 2) 88g CO23) 1 mol CO2 4) 33.6 lit CO2

5. A system is provided with 50 Joules of heatand the work done on the system is 10 Joules.What is the change in internal energy of thesystem in Joules ?1) 60 2) 40 3) 50 4) 10

Heat Capacity and Specific Heat6. The for inert gases is

1) 1.33 2) 1.66 3) 2.13 4) 1.997. The molar heat capacity of water is

1) 4.184 JK-1 2) 75.3 JK-1

3) 185 JK-1 4) 1 JK-1

Measurement of H And E

8. For a reaction 2 2 2A s B g C g D g .If heat of reaction at constant pressure is -28K.cal. The heat of reaction at constant volumeat 270C is1) -27.4 K.cal 2) 27.4 K.cal3) 28.4 K.cal 4) 28 K.cal

9. A sample of CH4 of 0.08g was subjected to

combustion at 270C in a bomb calorimeter. Thetemperature of the calorimeter system wasfound to be raised by 0.250C. If heat capacityof calorimeter is 18KJ, H for combustion ofCH4 at 270C is1) -900 KJ/mole 2) -905 KJ/mole3) -895KJ/mole 4) -890KJ/mole

10. 100ml of water at 200C and 100ml of water at400C are mixed in calorimeter until constanttemperature reached. Now temperature of themixture is 280C. Water equivalent ofcalorimeter is1) 50J 2) 104.5 J 3) -24.2J 4) 209J

11. Heat of combustion of benzoic acid (C6H5COOH) at constant volume at 250C is -3233KJ/mole. When 0.5g of benzoic acid isburnt in bomb calorimeter, the temperature ofcalorimeter increased by 0.530C. Now in thesame bomb calorimeter 1g of C2H6 burnt thentemperature increased by 2.040C. H forcombustion of C2H6 is1) -1530KJ/mole 2) -1536.2 KJ/mole3) -1522.8KJ/mole 4) +1536.2KJ/mole

Exothermic and Endothermic Reactions12. Which of the following is an exothermic

reaction?

1) 2 2 180.8 2g g gN O kJ NO

2) 2 2 33 92 . 2g g gN H k J NH

3) 2 2 131.3graphite g g gC H O CO H kJ

4) 22 91.9 .graphite s lC S CS k J

13. Which of the following is an exothermicreaction?

1) 2 2 2 ; 184.6g g gH Cl HCl H KJ

2) 2 2 2 ; 180.8g g gN O NO H KJ

3) 2 2 2 131.4graphite g gC H O CO H KJ

4) 22 91.9graphite g lC S KJ CS

Enthalpy of Reaction(Enthalpy of Phase Transitions)

14. The H for the conversion of C(diamond) to C(graphite) when the following reactions are given

9THERMODYNAMICS



C(diamond)+O2(g)CO2 (g); H = -94.5 K.cal.C(graphite)+O2(g) CO2(g); H = -94.0 K.cal.1) -188.5 K.cal 2) +188.5 K.cal3) +0.5 K.cal 4) -0.5 K.cal

15. H;gHI2sIgH 22 = 51.9 KJ/mole

2 2 2 ;H g I g HI g H = -9.2 KJ/mole.

The heat of reaction of gIsI 22 1) -61.1 KJ 2) +61.1 KJ3) +31.1 KJ 4) 50.1 KJ

16. If heats of combustion of red phosphorus andyellow phosphors are - 8.78KJ and - 9.19 Kjrespectiely then heat of transition of yellowphosphorus to red phosphrus will be1)17.97 KJ 2) -17.97 Kj3) -1.13 Kj 4) +1.13KJ

Enthalpy of Formation17. The heats of combustion for C, H2 and CH4

are -349, -241.8 and -906.7 KJ respectively.The heat of formation of CH4 is1) 174.1 KJ 2) 274.1 KJ3) 374.1 KJ 4) 74.1 KJ

18. From the thermo chemical reactions,

.KJ5.110H;COO21C 2graphite

.KJ2.283H;COO21CO 22 H for

the reaction, 22graphite COOC is1) -393.7 KJ 2) +393.7 KJ3) -172.7 KJ 4) +172.7 KJ

19. The heats of combustion of carbon, hydrogenand ethane are 94, 68.3 and 373 Kcal/molerespectively. The heat of formation of ethanein K.cals/mole is1) +19.9 2) -19.9 3) +39.8 4) -39.8

20. If the heats of formation of C2H2 and C6H6are 230 KJ mol-1 and 85 KJ mol-1 respectively,the H value for the trimerisation of C2H2 is1) -605 KJ 2) -205 KJ3) 205 KJ 4) 605 KJ

01) 2 02) 3 03) 1 04) 2 05) 1 06) 207) 2 08) 1 09) 2 10) 4 11) 2 12) 213) 1 14) 4 15) 2 16) 3 17) 4 18) 119) 2 20) 1

1.2

1

2.303 logVW nRTV

2. W P V 3. U q w 4. 22 gm U

x? 4U 5. U q w

8. H U nRT 9.MU ZW

H U nRT

10. 2 2 3

13 1

m t tW m

t t

11.MU ZW

H U nRT

14. Graphite DiamondH H H

15. 2 2I g I SH H H 16. Red P Yellow PH H H

17. R PH H H 2 42C H CH

18. P RH H H

19. 2 2 62 3C H C H R PH H H

20. 2 2 6 63C H C H P RH H H

.

10 THERMODYNAMICS



Internal Energy, Enthalpy, Work Done AndFirst Law Of Thermo Dynamics

1. The enthalpy change involved in the oxidationof glucose is -2880 KJ mol-1 25% of thisenergy is available for muscular work. If 100KJ of muscular work is needed to walk onekilometer, what is the maximum distance thata person will be able to walk after consuming120 gm of glucose?1) 4.8 km 2) 9.6 km 3) 2.4 km 4) 7.2 km

2. Three statements are given belowi) The enthalpy of any element is zero in theirstandard stateii) The heat of neutralisation for any strongacid and strong base at 25ºC is -13.7 kJ/moleiii) Q E W is a mathematical form of firstlaw of thermodynamics if work is done by thesystem according IUPAC1) i only correct 2) ii and iii are correct3) i and iii are correct 4) all are correct

3. I: 2 sH O II: 2 lH O III: 2 gH OFor the above compounds the correct ofinternal energy content is1) I II III 2) I II III 3) I II III 4) III II I

4. A gas contained a cylinder fitted with a frictionless piston expands against a constantpressure 1atm from a volume of 4litre tovolume of 14litre. In doing so, it absorbs 800Jthermal energy from surrounding. then the

E for the process is1) - 213.7 J 2) - 112 J 3) - 50J 4) - 25 J

5. The amount of work done by 2mole of an idealgas at 298K in reversible isothermal expansionfrom 10litre to 20litre is1) -120J 2) - 2452J 3) -3434.9J 4) 2200J

6. 5moles of an ideal gas at 270C expandsisothermally and reversibly from a volume of6L to 60L. The work done in KJ is1) –14.7 2) -28.72 3) +28.72 4) –56.72

7. 1mole of a gas is heated at constant pressureto raise its temperature by 10C. The workdone in Joules is

1) -4.3 2) -8.3143) -16.62 4) Unpredictable

8. 3.0 moles of ideal gas is heated at constantpressure from 270C to 1270C. then the workexpansion of gas is1) - 2.494KJ 2) + 2.494KJ3) - 10.5KJ 4) + 10.5KJ

9. An ideal gas occupying a volume of 2 dm3and apressure of 5 bar undergoes isothermal andirreversible expansion against externalpressure of 1 bar. The final volume of the systemand the work involved in the process is1) 10 dm3, 1000 J 2) 8 dm3, –800 J3) 10 dm3, –800 J 4) 10 m3, –1000 J

10. A position filled with 0.04 mol of an ideal gasexpands reversibly from 50.0 mL to 375 mL ata constant temperature of 37.00C . As it doesso, it absorbs 208J of heat. The values of qand w for the process will be:(R=8.314J/molK)(ln 7.5=2.01) (AIEEE-2013)1) q = +208 J, w = +208 J2) q = +208 J, w = -208 J3) q = -208 J, w = -208 J4) q = -208 J, w = +208 J

11. A heat engine absorbs heat Q1 at temperatureT1and heat Q2 at temperature T2. Work doneby the engine is (Q1+Q2). This data1) Violates 1st law of thermodynamics2) Violates 1st law of thermodynamics if 1Q is -ve3) Violates 1st law of thermodynamics if 2Q is -ve4) Does not violate 1st law of thermodynamics

12. H U for the formation ofcarbonmonoxide from its elements at 298K is1) -2477.57 J mol-1 2) 2477.57 J mol-13) -1238.78 J mol-1 4) 1238.78 J mol-1

13. The gas in a refrigerator causes cooling onexpansion becuase:1) Work done by the gas is converted into heat2) Heat of the gas is lost as work is done by thegas3) The heat is spread over a large space4) None of the above

Heat Capacity and Specific Heat14. 5moles of oxygen are heated at constant

volume from 010 C to 020 C . the change ininternal energy of a gas.

11THERMODYNAMICS



1 1 1 17.03 deg 8.31 degPC calmol and R Jmol 1) 125cal 2) 252cal 3) 50cal 4) 500cal

15. The molar heat capacity of water is 75 JK-1

mole-1. What is the amount of heat required toraise the temperature of 100g of water from300K to 302.4 K ?1) 10J 2) 1000 J 3) 375 J 4) 10 J

01) 1 02) 3 03) 4 04) 1 05) 3 06) 207) 2 08) 1 09) 3 10) 2 11) 4 12) 413) 2 14) 2 15) 2

1.2880 25 120180 100 100

2. The heat of neutralization for any strong acid andstrong base is -13.7 k Cal/mole at 250C

3. Gas > liquid > solid

4 to 6. 2

1

12.303 log VW nRTV

8. W nR T 7, 9. W P V 10. U Q W 11. First law of thermodynamics12. H U nRT 13. Heat of the gas is lost as work is done by the gas14. P VC C R ; Vq C T

Since W =0 U q W ; U q

15. PqCT

1. Thermodynamics is not concerned about1) energy changes involved in a chemical reaction2) the externt to which a chemical reaction proceeds3) the rate at which a reaction proceeds4) the feasibility of chemical reaction

2. Which of the following statements is correct?

1) The presence of reacting species in a coveredbeaker is an example of open system2) There is an exchange of energy as well as matterbetween the system and the surroundings in aclosed system3) The presence of reactants in a closed vesselmade up of copper is an example of closed system4) The presence of reactants in a thermos flask orany other closed insulated vessel is an example ofa closed system

3. The state of gas can be described by quotingthe relationship between1) pressure ,volume,temperature2) temperature,amount , pressure3) amount ,volume,temperature4) pressure ,volume,temperture ,amount

4. The volume of gas is reduced to half from itsoriginal volume .The specific heat will1) reduce to half 2) be doubled3) remain constant 4) increase four times

5. During complete combustion of one mole ofbutane, 2658KJ of heat is released. Thethermochemical reaction for above change is

1) 4 210 2 22 13 8 10g g g lC H O CO H O

12658.0cH kJ mol

2) 4 210 2 213 4 52g g g lC H O CO H O

11329.0cH kJ mol

3) 4 210 2 213 4 52g g g lC H O CO H O

12658.0cH kJ mol

4) 4 210 2 213 4 52g g g lC H O CO H O

12658.0cH kJ mol

6. 0fU of combustion of 4 gCH at certain

temperature is -393kJ mol–1. The value of0

f H is

1) zero 2) 0fU

3) 0fU 4) equal to 0

fU

12 THERMODYNAMICS



7. In an adiabatic process, no tranfer of heattakes place between system and surroundings.Choose the correct option for free expansionof an ideal gas under adibatic condition fromthe following.1) 0, 0, 0q T w

2) 0, 0, 0q T w

3) 0, 0, 0q T w

4) 0, 0, 0q T w 8. The pressure-volume work for an ideal gas can

be calculated by using the expressionf

i

V

exV

w p dV . The work can also be

calculated form the PV plot by using the areaunder curve within the specified limits. Whenan ideal gas is compressed 1) reversibly or 2)irreversibly from V1 to Vf, choose the correctoption1) w (reversible ) = w (irreversible)2) w (reversible )< w (irreversible)3) w (reversible ) > w (irreversible)4) w (reversible)= w (irreversible) + .exp V

9. The entropy change can be calculated by using

the expression revqST

when water freezes

in a glass beaker choose the correct statementamong the following

1) systemS decreases but surroundingsS remainsthe same

2) systemS increases but surroundingsS decreases

3) systemS decreases but surroundingsS increases

4) systemS decreases but surroundingsS also

decrease10. On the basis of thermochemical equation (i),(ii)

and (iii) ,find out which of the algebric relationships given in options (1) to (4) is correct

i) 1

2 2 ; rgraphite g gC O CO H x kJ mol

ii) 1

21 ;2 rgraphite g gC O CO H y kJ mol

iii) 1

2 21 ;2 rg gCO O CO H z kJ mol

1) z=x+y 2) x = y = z3) x=y+z 4) y = 2z - x

11. Consider the reactions given below .On thebasis of these reactions find out which of thealgerbric relations given in options (1) to (4) iscorrect ?i) 1

44 ; rC g H g CH g H x kJ mol

2 4, 2 ;C graphite s H g CH g

ii) 1r H y kJ mole

1) x = y 2) x = 2y 3) x > y 4) x < y12. The enthalpies of elements in their standard

states are taken as zero .The enthalpy offormation of a compound1) is always negative2) is always positive3) may be positive or negative4) is never negative

13. Enthalpy of sublimation of a substance is equal to1) enthalpy of fusion +enthalpy of vapourisation2) enthalpy of fusion3) enthalpy of vapourisation4) twice the enthalpy of vapourisation

14. Which of the following is not correct?1) G is zero for a reversible reaction2) G is positive for a spontaneous reaction3) G is negative for a spontaneous reaction4) G is positive for a non-spontaneous reaction

MORE THAN ONE OPTIONCORRECT15. Thermodynamics mainly deals with

1) interrelation of various forms of energy and theirtransformation from one form to another2) energy changes in the processes which dependsonly on initial and final states of the microscopoicsystems containing a few molecules3) how and at what rate these energytransformations are carried out4) the system in equilibrium state or moving fromone equilibrium state to another equilibrium state

1) C 2) C 3) D 4) C 5) C 6) B7) C 8) B 9) C 10) C 11)C 12) C13) A 14) B 15) A,D

13THERMODYNAMICS



1. Thermodynamics in not concerned with rate atwhich a reaction proceeds .The rate of reaction isdealt by kinetics

2. For a closed vessel made up of copper, no mattercan be exchanged between the system and thesurrounding but energy exchange can occur throughits walls

3. State of a gas can be described by state functionor state variables which are pressure, volume,temperature and amount of the gas(PV= nRT)

4. The specific heat of a substance is the heat requiredto raise the temperature of 1 gram of a substance byone degree (1K or 10 C ). It is an intensive propertyand is indepedent of the voulme of the substance

5. Exothermic reaction for combustion of one moleof butane is representend as:

4 210 2 213 4 52g g gC H O CO H O

12658.0cH kJ mol

6. 0 0f f gH U n RT

7. For free expansion w = 0For adiabatic process q = 0From first law of thermodynamics,

0U q w Since there is no change of internal energy, hencetemperature will also remain constant ,i.e., 0T

8. w (reversible) < w (irreversible)Area under the curve is greater in irreversiblecompression than that of reversible compression

Pres

sur e

fViV

Reversible compressionVolume

Pres

sure

fViV

Irreversible compressionVolume

9. During the process of freezing enegry is released,which is abosorbed by the surroundings

;revsystem

qST

rev

surroundingsqST

Therefore the entropy of the system decreases andthat of surroundings increases

11. x > y because same bonds are formed in reactions(i) and (ii) but bonds between reactant moleculesare broken ,only in reactions (ii). As energy isabsorbed when bonds are broken energy releasedin reaction (i) is greater than that in reaction (ii)

12. Heat of formation of a compound may be positiveor negative.

13. Since sublimation =fusion+Vapourisation15. Thermodynamics deals with interrelation of various

forms of energy and their tranformation into eachother. It also deals with thermal or mechancialequilbinun .However ,it does not tell any thing aboutthe rate of reaction

Thermo Dynamical Processes1. Boiling water in a closed steel tank is an

example of1) Closed system 2) Insulated system3) Open system 4) Adiabatic system

2. Which parameter is not constant in anadiabatic Process1) Temperature 2) Enthalpy3) Internal energy 4) 1 and 3

3. In isothermal process if heat is evolved from

14 THERMODYNAMICS



the system then1) Internal energy remains constant2) Change in internal energy is zero3) Change in entropy is zero4) Change in free energy is zero

Internal Energy, Enthalpy and Work done& Ist Law:

4. One mole of an ideal gas is allowed to expandreversibly and adiabatically from temperatureof 027 C . If the work done during the processis 3KJ, the final temperature will be equal to

20vC JK

1) 100K 2) 150K 3) 295 4) 026.85 C5. Work done on a system when one mole of an

an ideal gas at 500 K is compressedisothermally and reversibly to 1/10th of itsoriginal volume. (R = 2cal)1) 1 Kcal 2) 2.303 Kcal3) 4.606 Kcal 4) 2.303 cal

6. 2.8g of 2N gas at 300 K and 20 atm wasallowed to expand isothermally against aconstant external pressure of 1 atm. CalculateW for the gas.1) + 236.95 J 2) + 136.95 J3) - 236.95 J 4) + 136.95 J

7. When a sample of gas expands from 4.0L to12.0 L against a constant pressure of 0.30 atm,the work involved is1) 243.19 J 2) -243.19 J3) 234.19 J 4) -234.19 J

8. 5 mol of gas at 5 atmospheric pressurecontained in a 100 L cylinder absorbed 30.26Kj of heat when it expanded to 200 L at 2atmospheric pressure. The change in theinternal energy of gas is1) +20.26J 2) +20.26Kj3) -20.26 KJ 4) 10Kj

9. Frictionless and weightless piston was fittedinto a cylinder containing a gas This gas wasallowed to expand from one litre to 5 litreagainst a constant pressure of one atmospherein doing so, 200 J of heat was absorbed fromthe surrounding . The change in the internalenergy of the system is:1) +205.2 J 2) +205.2 KJ

3) -205.2 J 4) -405.2J10. A given mass of gas expands from the state A

to the state b by three paths 1, 2 and 3 asshown in the figure. If 1 2w ,w and 3wrespectively be the work done by the gasalong three paths then.

A

B

P

V

23

1) 1 2 3w w w 2) 1 2 3w w w

3) 1 2 3w w w 4) 2 3 1w w w

First Law of Thermodynamics11. One mole of liquid water at its boiling point

vaporises against a constant externalpressure of 1 atm.at the same temperature .Assuming ideal behaviour and initial volumeof water vapours are zero , the work done bythe system nearly1) -3102 J 2) +3102 J 3) -4268 J 4) + 4268 J

Heat Capacity and Specific Heat12. Heat capacity of water is 18 cal-degree-1-mol-

1. The quantity of heat needed to risetemperature of 18g water by 0.20C is X cal.Then amount of CH4(g) to be burnt to produceX cal heat is

4 2 2 22 2 , 200 .CH O CO H O H K Cal

1) 1.810–3 mol 2) 3.610–5 mol3) 0.0288 g 4) 0.288 mg

13. Heat capacity (CV) of an ideal gas is X KJ/mole/K. To rise its temperature from 298K to318K, heat to be supplied per 10g gas will be(in KJ) [MW=16]1) 16X 2) 6.25X 3) 32X 4) 12.5X

Mesurement Of H And E14. The relationship between H and E for the

reaction gPClgClgPCl 523 is givenas1) RTEH 2) RTEH

15THERMODYNAMICS



3) RT2EH 4) RT2EH 15. If E is the heat of reaction for

(1)52 22(g) 2(g) (1)C H OH 3O 2CO 3H O atconstant volume, the H (Heat of reaction atconstant pressure) at constant temperature is1) RT2EH 2) RT2EH 3) RTEH 4) RTEH

01) 1 02) 4 03) 12 04) 2 05) 2 06) 307) 2 08) 4 09) 3 10) 2 11) 1 12) 413) 4 14) 2 15) 4

1. Closed vessel2. In adiabatic process work is done at the cost of

internal energy

3. U f T for ideal gas

4. vqCT

For adiabat ic Process q=0,

E W ;

VU W nC T

3000 1 20 .J J T T2 = 150 K

5.2

1

2.303 log VW nRTV

6. 2 1irr exW P V V

2 1ex

nRT nRTPP P

2 1

1 1exP nRT

P P

7. W P V 8. U q W

9. U q W 10. More the area under the curve, more is the work

done

11. 2 12

0ex exnRTW P V V PP

12.MU Zx xW

; H U nRT ;Q = mst

13.10 20 12.516V Vq nC T x x

14. 1 2 1n 15. 2 3 1n

.

Internal Energy, Enthalpy and Work Done,First Law

1. A gas is allowed to expand reversibly underadiabatic conditions. What is zero for such aprocess ?1) G 2) T 3) S 4) None of these

Heat Capacity and Specific Heat2. One mole of an ideal diatomic gas absorbs

1831.4J of heat and does 1000J of work at270C. What is the change in temperatureobserved ?1) 250C 2) 400C 3) 370C 4) 670C

3. Calculate the change in enthalpy for hefollowing process at 1 atm

0 02 2,50 ,150H O l C H O g C

given that vH at 1000C is 40.7 kJ mol-1

1 12 , 75.0pC H O l Jmol K

1 12 , 33.3pC H O g Jmol K

1) 64.8 kJ 2) 52.4 kJ3) 48.6 kJ 4) 46.1kJ

Measurement of H and E4. 16 g of O2 at 280C is compressed to half of its

initial volume under reversible isothermalconditions. If this gas behaves ideally, theworkdone and change in internal energy in thisprocess are1) U 0 and w=-1247.1 ln 2 J2) U 0 and w = +1247.1 ln 2J

16 THERMODYNAMICS



3) w = U = +1247.1 ln 2J4) U -1247.1 ln 2J andw = +1247.1 ln 2J

5. In the following process/es H U

(A) 2 2C s O g CO g

(B) 2 212

CO g O g CO g

(C) 2 2 33 2N g H g NH g

(D) 3 2CaCO s CaO s CO g

The correct answer is1) A,B 2) B,C,D 3) A,B,C 4) B,C

Exothermic and Endothermic Reactions6. Which of the following will involve evolution

of heat ?1) Dissolution of conc. H2SO4 in water2) Formation of NO in atmosphere3) Conversion of molecular hydrogen to atomichydrogen4) Formation of water gas from coal and steam

Enthalpy of Reaction(Enthalpy of Phase Transitions)

7. The sublimation energy of I2 (solid) is 57.3 KJ/mole and enthalpy of fusion is 15.5 KJ/mole.The enthalpy of vapourisation of I2 is1) 48.1 KJ/mole 2) -48.1 KJ/mole3) 72.8 KJ/mole 4) -72.8 KJ/mole

Enthalpy of Formation8. Given the enthalpy of formation of CO2 (g) is

–94.0 KJ, of CaO (s) is -152 KJ, and theenthalpy of the reaction CaCO3 (s) CaO(s) + CO2 (g) is 42 KJ, the enthalpy offormation of CaCO3 (s) is1) -268 KJ 2) +202 KJ3) -202 KJ 4) -288 KJ

9. ;cal.K94H;COOC 22

.cal.K7.67H;COO21CO 22 On the

basis of the above data, the heat of formationof ‘CO’ is1) -26.3 K.cals 2) +26.3 K.cals3) +52.6 K.cals 4) +13.2 K.cals

Hess Law10. What is the heat of formation of C6H6, given

that the heats of combustion of Benzene,carbon and Hydrogen are 782, 94 and 68 K.Calrespectively1) +14 K.Cal 2) -14 K.Cal3) +28 K.Cal 4) -28 K.Cal

11. How much energy is released when 6 mole ofoctane is burnt in ? Given that the heat offormation of CO2, H2O and C8H18 respectivelyare -390, -240, and +160 KJ/mole.1) -32.6 MJ 2) -37.4 MJ3) -35.5 MJ 4) -20.0 MJ

Enthalpy of Combustion12. The heat of combustion of methane is -880 KJ

mol-1. The quantity of heat liberated in thecombustion of 3.2 g methane is1) -88 KJ 2) +88 KJ3) +176 KJ 4) -176 KJ

13. When 6 g carbon is burnt in a sufficient amountof oxygen, the heat evolved is x KJ. The heatof combustion of carbon is1) –x KJ 2) –2x KJ3) –4x KJ 4) –8x KJ

14. The heats of combustion of C2H4 , C2H6 and H2gases are –1409.5 KJ, –1558.3 KJ and –285.6KJ respectively. The heat of hydrogenation ofethene is1) -136.8 KJ 2) -13.68 KJ3) 273.6 KJ 4) 1.368 KJ

1) 3 2) 2 3) 4 4) 2 5) 4 16) 17) 1 8) 4 9) 1 10) 1 11) 1 12) 313) 2 14) 1

1. 0 0qq ST

2. V Vq C T

831.452

V

V

q JTC R

17THERMODYNAMICS



3. 0 0 02 2 20 50 100 150l l gH C H O C H O C

Total heat absorbed

= . 50 100temp rise H

. 100 150vap temp riseH H

P Vap PC T H C T

4. U Q W 5. H U nRT 6. Dissolution of sulphuric acid in water is exothermic

7. 2 2 2I S I l I g

SE Fusion EvaH H H

8. 3 2CaCO CaO CO ; P RH H H

9. 212

C O CO ; P RH H H

10. 2 6 66 3C H C H ; P RH H H

11. 8 18 2 2 225 8 92

C H O CO H O

P RH H H 12. 16 gm -880 3.2 gm ?13. 6 gm x 12 gm ?14. 2 4 2 2 6C H H C H P RH H H

CHANAKYA ACADEMY OF PROFESSIONAL STUDIES Jr|11th IIT...

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